Finding the Magnitude and Angle of a Problem: Two Solutions

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davidwinth
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Homework Statement
Find the required force and angle for equilibrium
Relevant Equations
Included Sum of Forces equals zero
Hello,

I solved this problem two ways and got the same value for the magnitude but different values for the angle. I am wondering which is correct. I showed my work, so I hope someone can tell me if one method is just invalid.
t7EJy2m.jpg
 
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mjc123 said:
sinθ = x has two solutions, as sinθ = sin(180°-θ). Use the cos rule to determine the sign of cos(φ+45°).
Hmm. I am not sure this answers the question. The fact that I got the same value for the magnitude with both approaches means the angles cannot be different, right? I am physically speaking here, whatever the properties of the sine function. One of them must be wrong - they are incompatible. I just don't know how to tell which one is wrong.
 
davidwinth said:
Hmm. I am not sure this answers the question. The fact that I got the same value for the magnitude with both approaches means the angles cannot be different, right? I am physically speaking here, whatever the properties of the sine function. One of them must be wrong - they are incompatible. I just don't know how to tell which one is wrong.
Your problem is that there are two solutions to ##sin(\phi+45 deg)=850/867.28##. One is a bit under 90 degrees, the other a bit over.
Geometrically, this corresponds to the fact that knowing two sides of a triangle and the non-included angle is not sufficient to define the triangle.
 
davidwinth said:
Hmm. I am not sure this answers the question. The fact that I got the same value for the magnitude with both approaches means the angles cannot be different, right?
Right. But to make it clear, you calculated the wrong angle in your first method (as others have indicated).
##\frac {850}{sin(\phi + 45º)} = \frac {433.64}{sin(30º)}##
gives
##\phi + 45º = 78.5º## OR ##101.5º## (since sin(x) = sin(180º - x))

But you picked the wrong one! In fact you can tell from your diagram that ##\phi + 45º## must be more than 90º.

If you’re still not clear about this, try the video:
 
Steve4Physics said:
Right. But to make it clear, you calculated the wrong angle in your first method (as others have indicated).
##\frac {850}{sin(\phi + 45º)} = \frac {433.64}{sin(30º)}##
gives
##\phi + 45º = 78.5º## OR ##101.5º## (since sin(x) = sin(180º - x))

But you picked the wrong one! In fact you can tell from your diagram that ##\phi + 45º## must be more than 90º.

If you’re still not clear about this, try the video:

Ah, so the first approach is the wrong one. To clarify: My drawing is not to scale so I was not going to decide which angle to pick based on that! Also, the problem statement indicated that the angle phi should be less than 45 degrees. If I had used only the first approach, there seems no way to know which solution is correct except to go by the problem statement (which is wrong, apparently)! The second approach is therefore more robust.

That was my question: how would someone who solved the problem with the approach know that one of the solutions is impossible without checking by the second approach?

People mentioned I should use, "The cos rule" and I am not sure what they mean. When I google this, the law of cosines shows up, which I did use in the first approach already.

Thanks
 
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“Ah, so the first approach is the wrong one. “

Not wrong - but ambiguous. In this particular situation using the sine rule gives ambiguous values for ##\phi + 45^0## So there is some extra thinking needed to decide which of the 2 possible values to use.
___________

“My drawing is not to scale so I was not going to decide which angle to pick based on that!”

Fair enough. But it’s not too bad a drawing. If you were aware of the ambiguity-risk, you might get a hint from the drawing.
____________

“Also, the problem statement indicated that the angle phi should be less than 45 degrees.”

You didn’t tell us that! Either the problem statement is incorrect or your interpretation of the question is incorrect (e.g. your diagram could be incorrect). But there is no way we can tell from the information you have provided. We’re assuming your diagram is correct.
____________

“If I had used only the first approach, there seems no way to know which solution is correct except to go by the problem statement (which is wrong, apparently)! The second approach is therefore more robust..”

You can easily tell which of the ambiguous solutions from the 1st approach is correct. You could consider the projection of ##\vec F## onto ##\vec B##.

##850cos(30^0) = 736##. This is bigger than ##650##, so ##\phi + 45^0## must be bigger tan ##90^0##.

The first approach is not bad as long as you recognise and correctly address the ambiguity issue.
_____________

“That was my question: how would someone who solved the problem with the approach know that one of the solutions is impossible without checking by the second approach? “

See previous point.
_____________

“People mentioned I should use, "The cos rule" and I am not sure what they mean. When I google this, the law of cosines shows up, which I did use in the first approach already. “

You can use the cosine rule for a second time in your 1st approach (instead of using the sine rule) because you know all 3 sides:
##850^2 = 433.64^2 + 650^2 - 2 \times 443.64 \times 650cos(\phi + 45^0)##

This gives the same answer as your second method.